The image presents a complex number $z = 3 - 3i\sqrt{3}$ and asks to find $\sqrt[4]{z^2}$, the fourth root of the square of $z$.

Let's break it down:

1. Step 1: Calculate $z^2$
   We need to square the complex number $z = 3 - 3i\sqrt{3}$.

   $z^2 = (3 - 3i\sqrt{3})^2$

   Expanding this expression:

   $z^2 = 3^2 - 2(3)(3i\sqrt{3}) + (3i\sqrt{3})^2$ $z^2 = 9 - 18i\sqrt{3} + (9)(-3)$ $z^2 = 9 - 18i\sqrt{3} - 27$ $z^2 = -18 - 18i\sqrt{3}$

2. Step 2: Find the fourth root of $z^2$
   We now need to find $\sqrt[4]{-18 - 18i\sqrt{3}}$. To do this, it's easier to convert the complex number into polar form.

   The complex number $-18 - 18i\sqrt{3}$ can be written as $r(\cos \theta + i \sin \theta)$, where:

   • $r = \sqrt{(-18)^2 + (-18\sqrt{3})^2} = \sqrt{324 + 972} = \sqrt{1296} = 36$,
   • The argument $\theta$ can be calculated as $\theta = \tan^{-1}\left(\frac{-18\sqrt{3}}{-18}\right) = \tan^{-1}(\sqrt{3}) = \frac{2\pi}{3}$.

   So, in polar form:

   $z^2 = 36 \left( \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) \right)$

3. Step 3: Apply the fourth root
   To find $\sqrt[4]{z^2}$, we take the fourth root of both the modulus and the argument:

   • The modulus becomes $\sqrt[4]{36} = \sqrt{6}$,
   • The argument becomes $\frac{1}{4} \times \frac{2\pi}{3} = \frac{\pi}{6}$.

   Therefore, $\sqrt[4]{z^2}$ in polar form is:

   $\sqrt[4]{z^2} = \sqrt{6} \left( \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \right)$

   Simplifying:

   $\sqrt[4]{z^2} = \sqrt{6} \left( \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \right)$ $\sqrt[4]{z^2} = \frac{\sqrt{18}}{2} + i\frac{\sqrt{6}}{2}$

   Hence, $\sqrt[4]{z^2}$ is:

   $\boxed{\frac{\sqrt{18}}{2} + i \frac{\sqrt{6}}{2}}$

Let me know if you would like a more detailed explanation or have any questions!

Here are some related questions:

1. How do you find the square of a complex number geometrically?
2. How do you convert a complex number into polar form?
3. What is the process of extracting the $n$-th root of a complex number?
4. What is the significance of Euler's formula in complex number operations?
5. Can you generalize the process of finding the fourth root of any complex number?

Tip: When working with complex numbers, converting to polar form can simplify powers and roots considerably.